Question 211575
rate * time = number of gallons filled
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r1 = hose 1 rate
t1 = hose 1 time
r2 = hose 2 rate
t2 = hose 2 time
g = number of gallons
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r1*t1 = g
t1 = 4
4*t1 = g
t1 = g/4
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r2*t2 = g
r2 = g/t2
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(r1+r2)*(t2-2) = g
substitute g/4 for r1 and g/t2 for r2 to get:
(g/4 + g/t2) * (t2-2) = g
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divide both sides of equation by g to get:
(1/4 + 1/t2) * (t2-2) = 1
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multiply both sides of the equation by 4*t2 to get:
(t2 + 4) * (t2 - 2) = 4*t2
multiply factors out to get:
(t2)^2 + 4*t2 - 2*t2 - 8 = 4*t2
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subtract 4*t2 from both sides of the equation and combine like terms to get:
(t2)^2 - 2*t2 - 8 = 0
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factor this to get:
(t2 - 4) * (t2 + 2) = 0
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t2 = 4 OR t2 = -2
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since t2 can't be negative, only good solution would be:
t2 = 4
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Answer is it takes pipe 2 4 hours to fill the tank.
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let g be any number of gallons.
suppose 100 gallons
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4r1 = 100
r1 = 25
4r2 = 100
r2 = 25
working together we get:
(r1+r2)*2 = 100
(50)*2 = 100
100 = 100
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checks out ok.
hose 1 takes 4 hours
hose 2 takes 4 hours
working together they take 2 hours
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